When the Heat gain or lose = the mass * specific heat * ΔT
and when we have the mass of gold coin= 40 g
and the specific Heat of gold= 0.13 J/g°
and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C
so by substitution:
∴Heat H = 40 g * 0.13 J/g° * -40
= - 208 J
Answer:
We need 1.1 grams of Mg
Explanation:
Step 1: Data given
Volume of water = 78 mL
Initial temperature = 29 °C
Final temperature = 78 °C
The standard heats of formation
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Step 2: The equation
The heat is produced by the following reaction:
Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)
Step 3: Calculate the mass of Mg needed
Using the standard heats of formation:
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg
(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required
(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg
We need 1.1 grams of Mg
Explanation:
Elements in the same group have same number of valence electrons. And we know, the elements which have same number of valence electrons, have similar physical and chemical properties. Hence, the elements in the same group have similar physical and chemical properties.
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